If - z 1- z 2- z 1- z 2- then z 1- z 2-A- 2 n -1 -2- n - ZB- n -4- n - ZC- 2 n -1 - n - ZD- 2 n - n - Z

Given that nbsp;|z_1+z_2|=|z_1 z_2| ⇒ |z_1+z_2|^2=|z_1 z_2|^2 [ [ ∵|z_1 z_2|^2=|z_1|^2+|z_2|^2 2Re(z_1z_2 ),; |z_1+z_2|^2=|z_1|^2+|z_2|^2+2Re(z_1z_2 ) ]] ⇒ ...

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Standard XII

Mathematics

Euler's Representation

If | z 1+ z 2...

Question

If $| z_{1} + z_{2} | = | z_{1} - z_{2} |$ then $arg z_{1} - arg z_{2} =$

(2 n + 1) \frac{π}{2}, n \in Z

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n π + \frac{π}{4}, n \in Z

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2 n π, n \in Z

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(2 n + 1) π, n \in Z

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Solution

The correct option is A $(2 n + 1) \frac{π}{2}, n \in Z$
Given that $| z_{1} + z_{2} | = | z_{1} - z_{2} |$
$\Rightarrow | z_{1} + z_{2} |^{2} = | z_{1} - z_{2} |^{2}$
$[\begin{matrix} ∵ | z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (z_{1}_{2}), | z_{1} + z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} + 2 R e (z_{1}_{2}) \end{matrix}]$
$\Rightarrow | z_{1} |^{2} + | z_{2} |^{2} + 2 R e (z_{1}_{2}) = | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (z_{1}_{2})$
$\Rightarrow 4 R e (z_{1}_{2}) = 0$
$\Rightarrow R e (z_{1}_{2}) = 0$

Let $z_{1} = r_{1} e^{i θ_{1}}$ and $z_{2} = r_{2} e^{i θ_{2}}$
Then $z_{1}_{2} = r_{1} r_{2} e^{i θ_{1}} e^{- i θ_{2}} = r_{1} r_{2} e^{i (θ_{1} - θ_{2})}$
$\Rightarrow z_{1}_{2} = r_{1} r_{2} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$
$∵ R e (z_{1}_{2}) = 0$
$∴ cos (θ_{1} - θ_{2}) = 0$
$\Rightarrow θ_{1} - θ_{2} = (2 n + 1) \frac{π}{2}, n \in Z$

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